Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 PisEmptyProof (⇔)
    ↳36 TRUE
  ↳37 QDP
    ↳38 UsableRulesProof (⇔)
    ↳39 QDP
    ↳40 QReductionProof (⇔)
    ↳41 QDP
    ↳42 Induction-Processor (⇐)
    ↳43 AND
      ↳44 QDP
        ↳45 DependencyGraphProof (⇔)
        ↳46 TRUE
      ↳47 QTRS
        ↳48 Overlay + Local Confluence (⇔)
        ↳49 QTRS
        ↳50 DependencyPairsProof (⇔)
        ↳51 QDP
        ↳52 DependencyGraphProof (⇔)
        ↳53 AND
          ↳54 QDP
            ↳55 UsableRulesProof (⇔)
            ↳56 QDP
            ↳57 QReductionProof (⇔)
            ↳58 QDP
            ↳59 QDPSizeChangeProof (⇔)
            ↳60 TRUE
          ↳61 QDP
            ↳62 UsableRulesProof (⇔)
            ↳63 QDP
            ↳64 QReductionProof (⇔)
            ↳65 QDP
            ↳66 QDPSizeChangeProof (⇔)
            ↳67 TRUE
          ↳68 QDP
            ↳69 UsableRulesProof (⇔)
            ↳70 QDP
            ↳71 QReductionProof (⇔)
            ↳72 QDP
            ↳73 QDPSizeChangeProof (⇔)
            ↳74 TRUE
          ↳75 QDP
            ↳76 UsableRulesProof (⇔)
            ↳77 QDP
            ↳78 QReductionProof (⇔)
            ↳79 QDP
            ↳80 QDPOrderProof (⇔)
            ↳81 QDP
            ↳82 PisEmptyProof (⇔)
            ↳83 TRUE
          ↳84 QDP
            ↳85 UsableRulesProof (⇔)
            ↳86 QDP
            ↳87 QReductionProof (⇔)
            ↳88 QDP
            ↳89 QDPSizeChangeProof (⇔)
            ↳90 TRUE
          ↳91 QDP
            ↳92 UsableRulesProof (⇔)
            ↳93 QDP
            ↳94 QReductionProof (⇔)
            ↳95 QDP
            ↳96 QDPSizeChangeProof (⇔)
            ↳97 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOG(x) → LOGITER(x, 0)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
LOGITER(x, y) → LE(s(0), x)
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → QUOT(x, s(s(0)))
LOGITER(x, y) → INC(y)
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x)) → INC(x)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x)) → INC(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • INC(s(x)) → INC(x)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LE(s(x), s(y)) → LE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(x), s(y)) → MINUS(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(27) TRUE

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUOT(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

(34) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(35) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(36) TRUE

(37) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(38) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(40) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

We have to consider all minimal (P,Q,R)-chains.

(42) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF(x1, x2, x3, x4)) = x1 + 2·x2 + x3   
POL(LOGITER(x1, x2)) = 3   
POL(false) = 0   
POL(inc(x1)) = 0   
POL(le(x1, x2)) = 1   
POL(minus(x1, x2)) = 2 + x1   
POL(quot(x1, x2)) = 0   
POL(s(x1)) = 2·x1   
POL(true) = 1   

At least one of these decreasing rules is always used after the deleted DP:
le(s(x), 0) → false


The following formula is valid:
z0:sort[a0],x:sort[a0].(((z0 =s(0)→le'(z0 , x )=true)∨(z0 =s2 (0)→le'(z0 , x )=true))∨false=true)


The transformed set:
le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

(43) Complex Obligation (AND)

(44) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

We have to consider all minimal (P,Q,R)-chains.

(45) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(46) TRUE

(47) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

Q is empty.

(48) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(49) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(50) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(51) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE'(s(x5), s(y3)) → LE'(x5, y3)
LE(s(x5), s(y3)) → LE(x5, y3)
QUOT(s(x18), s(y15)) → QUOT(minus(x18, y15), s(y15))
QUOT(s(x18), s(y15)) → MINUS(x18, y15)
INC(s(x25)) → INC(x25)
MINUS(s(x45), s(y36)) → MINUS(x45, y36)
EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(52) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 1 less node.

(53) Complex Obligation (AND)

(54) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(55) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(56) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

R is empty.
The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(57) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(58) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(59) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
    The graph contains the following edges 1 > 1, 2 > 2

(60) TRUE

(61) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x45), s(y36)) → MINUS(x45, y36)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(62) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(63) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x45), s(y36)) → MINUS(x45, y36)

R is empty.
The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(64) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(65) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x45), s(y36)) → MINUS(x45, y36)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(66) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(x45), s(y36)) → MINUS(x45, y36)
    The graph contains the following edges 1 > 1, 2 > 2

(67) TRUE

(68) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x25)) → INC(x25)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(69) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(70) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x25)) → INC(x25)

R is empty.
The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(71) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(72) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x25)) → INC(x25)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(73) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • INC(s(x25)) → INC(x25)
    The graph contains the following edges 1 > 1

(74) TRUE

(75) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x18), s(y15)) → QUOT(minus(x18, y15), s(y15))

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(76) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(77) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x18), s(y15)) → QUOT(minus(x18, y15), s(y15))

The TRS R consists of the following rules:

minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
minus(0, s(x0)) → 0

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(78) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(79) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x18), s(y15)) → QUOT(minus(x18, y15), s(y15))

The TRS R consists of the following rules:

minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
minus(0, s(x0)) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
minus(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.

(80) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(s(x18), s(y15)) → QUOT(minus(x18, y15), s(y15))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUOT(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

minus(x38, 0) → x38
minus(0, s(x0)) → 0
minus(s(x45), s(y36)) → minus(x45, y36)

(81) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
minus(0, s(x0)) → 0

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
minus(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.

(82) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(83) TRUE

(84) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x5), s(y3)) → LE(x5, y3)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(85) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(86) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x5), s(y3)) → LE(x5, y3)

R is empty.
The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(87) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(88) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x5), s(y3)) → LE(x5, y3)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(89) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LE(s(x5), s(y3)) → LE(x5, y3)
    The graph contains the following edges 1 > 1, 2 > 2

(90) TRUE

(91) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE'(s(x5), s(y3)) → LE'(x5, y3)

The TRS R consists of the following rules:

le'(s(x), 0) → true
le'(s(x5), s(y3)) → le'(x5, y3)
le'(0, y42) → false
le(s(x), 0) → false
le(s(x5), s(y3)) → le(x5, y3)
quot(0, s(y9)) → 0
quot(s(x18), s(y15)) → s(quot(minus(x18, y15), s(y15)))
inc(s(x25)) → s(inc(x25))
inc(0) → s(0)
minus(x38, 0) → x38
minus(s(x45), s(y36)) → minus(x45, y36)
le(0, y42) → true
quot(0, 0) → 0
quot(s(x2), 0) → 0
minus(0, s(x0)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a32](witness_sort[a32], witness_sort[a32]) → true

The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(92) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(93) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE'(s(x5), s(y3)) → LE'(x5, y3)

R is empty.
The set Q consists of the following terms:

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

We have to consider all minimal (P,Q,R)-chains.

(94) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

le'(s(x0), 0)
le'(s(x0), s(x1))
le'(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
quot(0, 0)
quot(s(x0), 0)
minus(0, s(x0))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a32](witness_sort[a32], witness_sort[a32])

(95) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE'(s(x5), s(y3)) → LE'(x5, y3)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(96) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LE'(s(x5), s(y3)) → LE'(x5, y3)
    The graph contains the following edges 1 > 1, 2 > 2

(97) TRUE